Title: Strange Mandelbrot Set
Post by: Endemyon on December 12, 2013, 06:48:11 PM
Hello Everybody !
I tried a formula wich obtains interesting results:
http://www.fractalforums.com/index.php?action=gallery;sa=view;id=15275
It's obtained from iteration of
Zn+1=(Zn^3+c)*(Zn^2+c)
I assume it's possible to do every formula's of type:
Zn+1=(Zn^p+c)*(Zn^q+c) with p,q>1 integers
because the two parts of the equation are mandelbrot's formulas.
Maybe it's even possible to multiply it with other terms of type (Zn^k+c)
It's possible to obtained corresponding julia sets just like the normal way, but they're not totally symetric. It seems like a composition of the two sets !
Does anyone is familiar with this kind of thing and tell me more about it ?
Friendly, a fractal fellow
Edit: A corresponding Julia's set:
http://www.fractalforums.com/index.php?action=gallery;sa=view;id=15278
And it appears to me that i might have misplaced this topic, could a great asministrator deplace it to the appropriate area ?
I tried a formula wich obtains interesting results:
http://www.fractalforums.com/index.php?action=gallery;sa=view;id=15275
It's obtained from iteration of
Zn+1=(Zn^3+c)*(Zn^2+c)
I assume it's possible to do every formula's of type:
Zn+1=(Zn^p+c)*(Zn^q+c) with p,q>1 integers
because the two parts of the equation are mandelbrot's formulas.
Maybe it's even possible to multiply it with other terms of type (Zn^k+c)
It's possible to obtained corresponding julia sets just like the normal way, but they're not totally symetric. It seems like a composition of the two sets !
Does anyone is familiar with this kind of thing and tell me more about it ?
Friendly, a fractal fellow
Edit: A corresponding Julia's set:
http://www.fractalforums.com/index.php?action=gallery;sa=view;id=15278
And it appears to me that i might have misplaced this topic, could a great asministrator deplace it to the appropriate area ?
Title: Re: Strange Mandelbrot Set
Post by: cKleinhuis on March 03, 2014, 08:33:45 AM
hey there, nice meeting you as well, i move it to the math section, what you did there is a kind of hybrid fractal,
and yes, basically this is one form of hybridisation, combining existing formulas to new ones, you connected the two
z^2 and z^3 mandelbrot equation with the multiplication operation ;)
and yes, basically this is one form of hybridisation, combining existing formulas to new ones, you connected the two
z^2 and z^3 mandelbrot equation with the multiplication operation ;)
Title: Re: Strange Mandelbrot Set
Post by: Rychveldir on March 06, 2014, 11:58:15 AM
In your Formula
(z_n^3+c))
Is c a constant? And did you use
at all? Or was it something like
(z_n^3+c)+z_0)
or
(z_n^3+z_0))
I also like exploring generalized versions of the mandelbrot set using polynomials of the form
They can generate really interesting effects.
Is c a constant? And did you use
or
I also like exploring generalized versions of the mandelbrot set using polynomials of the form
They can generate really interesting effects.
Title: Re: Strange Mandelbrot Set
Post by: cKleinhuis on March 06, 2014, 02:33:42 PM
i think the "c" is just the z0 value in this case ;)
but anyways, all of the combinations should provide interesting results ;)
but anyways, all of the combinations should provide interesting results ;)
Title: Re: Strange Mandelbrot Set
Post by: element90 on March 07, 2014, 12:11:16 PM
The difficulty with the formula
z = (z^3 + c)(z^2 + c)
is that the critical points can't be easily determined except for the point at zero.
The formula expands to
z = z^5 +cz^3 + cz^2 + c^2
so the critical points are the solutions of
f'(z) = 5z^4 + 3cz^2 + 2cz = 0
so
z(5z^3 + 3cz^2 + c) = 0
which implies that there is a critical point at 0, the remaining 3 critical values have to calculated for each value of c (the location in the complex plane).
z = (z^3 + c)(z^2 + c)
is that the critical points can't be easily determined except for the point at zero.
The formula expands to
z = z^5 +cz^3 + cz^2 + c^2
so the critical points are the solutions of
f'(z) = 5z^4 + 3cz^2 + 2cz = 0
so
z(5z^3 + 3cz^2 + c) = 0
which implies that there is a critical point at 0, the remaining 3 critical values have to calculated for each value of c (the location in the complex plane).
Title: Re: Strange Mandelbrot Set
Post by: kram1032 on March 07, 2014, 12:27:42 PM
Since that's a Polynomial of degree four, you can still solve for the generic critical point, however the expressions are rather complex:
http://www.wolframalpha.com/input/?i=5z%5E4%2B3+c+z%5E2+%2B+2+c+z+%3D+0
http://www.wolframalpha.com/input/?i=5z%5E4%2B3+c+z%5E2+%2B+2+c+z+%3D+0
Title: Re: Strange Mandelbrot Set
Post by: element90 on March 07, 2014, 01:51:03 PM
For higher level polynomials from a programming point of view it is even easier, the coefficients can be used to produce the companion matrix and the eigenvalues are the roots, provided you have the routines to do it.. It just has to be done for each location. Implementing this in a script with a program such as Ultra Fractal won't be easy.
I looked at
z = (z^2 + c)(z^2 + c)
so
f'(z) = 4z^3 + 2cz = 0
z(4z^2 + 2c) = 0
which has three critical points 0, -sqrt(-c) and sqrt(-c) which can be plugged into a UF or Gnofract4d script which it turns out produces three identical pictures which isn't surprising as the pictures are properly formed M4 Mandelbrots.
I looked at
z = (z^2 + c)(z^2 + c)
so
f'(z) = 4z^3 + 2cz = 0
z(4z^2 + 2c) = 0
which has three critical points 0, -sqrt(-c) and sqrt(-c) which can be plugged into a UF or Gnofract4d script which it turns out produces three identical pictures which isn't surprising as the pictures are properly formed M4 Mandelbrots.
Title: Re: Strange Mandelbrot Set
Post by: element90 on March 07, 2014, 02:41:38 PM
The formula
z = (z^2 + c)(z^4 + c)
has somewhat easier critical points to determine, for the values see the code below in the Gnofract4d script.
The critical points are solutions of
f'(z) = 6z^5 + 4cz^3 + 2cz = 0
z(6z^4 + 4cz^2 + 2c) = 0
which gives zero and the square roots of the roots of a quadratic.
Critical point 0
(https://copy.com/hqSr3VYbJBuO)
https://copy.com/hqSr3VYbJBuO (https://copy.com/hqSr3VYbJBuO)
Critical point z = sqrt((-z0 + sqrt(z0*z0 - 24*z0))/12) and sqrt((-z0 + sqrt(z0*z0 - 24*z0))/12)
(https://copy.com/kqLZT9DwdGFJ)
https://copy.com/kqLZT9DwdGFJ (https://copy.com/kqLZT9DwdGFJ)
Critical point z = sqrt((-z0 - sqrt(z0*z0 - 24*z0))/12) and -sqrt((-z0 - sqrt(z0*z0 - 24*z0))/12)
(https://copy.com/UMOwFuqWj8ed)
https://copy.com/UMOwFuqWj8ed (https://copy.com/UMOwFuqWj8ed)
Combined
(https://copy.com/BGUDDqTrGMt7)
https://copy.com/BGUDDqTrGMt7 (https://copy.com/BGUDDqTrGMt7)
What usually happens is that areas in one critical point picture that lack Mandelbrot buds are provided with them from an other critical point picture. Oddly the areas at the top and the bottom have areas where this isn't so, assuming I haven't made a mistake.
z = (z^2 + c)(z^4 + c)
has somewhat easier critical points to determine, for the values see the code below in the Gnofract4d script.
The critical points are solutions of
f'(z) = 6z^5 + 4cz^3 + 2cz = 0
z(6z^4 + 4cz^2 + 2c) = 0
which gives zero and the square roots of the roots of a quadratic.
Code:
M2.M4 {
; M2 Mandelbrot multiplied by an M4 Mandelbrot
init:
; z = 0
z0 = #pixel
; z = sqrt((-z0 + sqrt(z0*z0 - 24*z0))/12)
; z = sqrt((-z0 - sqrt(z0*z0 - 24*z0))/12)
z = -sqrt((-z0 + sqrt(z0*z0 - 24*z0))/12)
; z = -sqrt((-z0 - sqrt(z0*z0 - 24*z0))/12)
loop:
z2 = z*z
z = (z2 + #pixel)*(z2*z2 + #pixel)
bailout:
@bailfunc(z) < @bailout
default:
float param bailout
default = 4.0
endparam
float func bailfunc
default = cmag
endfunc
}Critical point 0
(https://copy.com/hqSr3VYbJBuO)
https://copy.com/hqSr3VYbJBuO (https://copy.com/hqSr3VYbJBuO)
Critical point z = sqrt((-z0 + sqrt(z0*z0 - 24*z0))/12) and sqrt((-z0 + sqrt(z0*z0 - 24*z0))/12)
(https://copy.com/kqLZT9DwdGFJ)
https://copy.com/kqLZT9DwdGFJ (https://copy.com/kqLZT9DwdGFJ)
Critical point z = sqrt((-z0 - sqrt(z0*z0 - 24*z0))/12) and -sqrt((-z0 - sqrt(z0*z0 - 24*z0))/12)
(https://copy.com/UMOwFuqWj8ed)
https://copy.com/UMOwFuqWj8ed (https://copy.com/UMOwFuqWj8ed)
Combined
(https://copy.com/BGUDDqTrGMt7)
https://copy.com/BGUDDqTrGMt7 (https://copy.com/BGUDDqTrGMt7)
What usually happens is that areas in one critical point picture that lack Mandelbrot buds are provided with them from an other critical point picture. Oddly the areas at the top and the bottom have areas where this isn't so, assuming I haven't made a mistake.
Title: Re: Strange Mandelbrot Set
Post by: jdebord on March 10, 2014, 10:11:17 AM
There are analytical formulas for polynomials up to degree 4. See for instance:
http://dlmf.nist.gov/1.11#iii (http://dlmf.nist.gov/1.11#iii)
Since we are looking for the roots of the derivative, this applies for polynomials up to degree 5.
http://dlmf.nist.gov/1.11#iii (http://dlmf.nist.gov/1.11#iii)
Since we are looking for the roots of the derivative, this applies for polynomials up to degree 5.
Title: Re: Strange Mandelbrot Set
Post by: Endemyon on March 31, 2014, 11:21:38 PM
Hey guys, thanks for your answers !
I can't resist to show you this julia set, obtained by Zn+1=(Zn^2+c)*(Zn^3+c)*(Zn^4+c) with C=0,7937+i0,0904
http://www.fractalforums.com/index.php?action=gallery;sa=view;id=15802
I don't understand why you need to find the roots of critical points (if it's not to have a completely connex Julia set ) ?
It's possible to have an approximation for higher degrees with the newton's method, but I think you don't learn something new ^-^
I can't resist to show you this julia set, obtained by Zn+1=(Zn^2+c)*(Zn^3+c)*(Zn^4+c) with C=0,7937+i0,0904
http://www.fractalforums.com/index.php?action=gallery;sa=view;id=15802
I don't understand why you need to find the roots of critical points (if it's not to have a completely connex Julia set ) ?
It's possible to have an approximation for higher degrees with the newton's method, but I think you don't learn something new ^-^
Title: Re: Strange Mandelbrot Set
Post by: jdebord on April 01, 2014, 10:12:46 AM
The critical point is required only for the Mandelbrot set, not for the Julia set.
The Julia set iteration always start with the pixel coordinates.
The Julia set iteration always start with the pixel coordinates.
Title: Re: Strange Mandelbrot Set
Post by: Endemyon on April 01, 2014, 06:47:10 PM
Quote
The critical point is required only for the Mandelbrot set, not for the Julia set.
The Julia set iteration always start with the pixel coordinates.
The Julia set iteration always start with the pixel coordinates.
If you do a Mandelbrot set for the equation Zn+1=Zn^2+C (the classic). C is the point which parcours the pixel coordinates and Z0=0.
So you don't need a critical point :)
So that doesn't not explain me why it's needed, is it from the software you use?
Now, why I was talking about critical point for a julia set:
If you do a julia set for the equation Zn+1=Zn^2+C. C is the chosen constant and Z0 is the point which is the pixel coordinates.
If you try do a julia set where C is the critical point of mandelbrot set, you obtain a julia set where all converging parts are connected. ( for any n)
:D
Title: Re: Strange Mandelbrot Set
Post by: jdebord on April 02, 2014, 09:05:34 AM
If f(z) denotes the iterated formula, the critical points are the roots of the equation f'(z) = 0
For the classical Mandelbrot set, f(z) = z^2 + c and f'(z) = 2z.
The equation is 2z = 0, hence z = 0
So, z0 = 0 is a critical point (the only one, in this case).
For the classical Mandelbrot set, f(z) = z^2 + c and f'(z) = 2z.
The equation is 2z = 0, hence z = 0
So, z0 = 0 is a critical point (the only one, in this case).
Title: Re: Strange Mandelbrot Set
Post by: TheRedshiftRider on June 10, 2014, 05:48:45 PM
I tried something else, I tried to divide the two parts of the basic formula and the results are nice:
(http://i.imgur.com/8cDPKlA.png)
(z^10+c)/(z^2+c)
(http://i.imgur.com/qg4vr57.png)
(z^10+c)/(z^8+c)
(http://i.imgur.com/8cDPKlA.png)
(z^10+c)/(z^2+c)
(http://i.imgur.com/qg4vr57.png)
(z^10+c)/(z^8+c)
Title: Re: Strange Mandelbrot Set
Post by: jdebord on June 11, 2014, 10:07:54 AM
Nice pictures, especially the second one !
I have tried to find a formula for the derivatives, assuming p > q > 1 :
 = \frac{z^p + c}{z^q + c} )
![f'(z) = \frac{z^{q-1}[(p - q)z^p+(pz^{p-q}-q)c]}{(z^q + c)^2}](/cgi-bin/mimetex.cgi? f'(z) = \frac{z^{q-1}[(p - q)z^p+(pz^{p-q}-q)c]}{(z^q + c)^2} )
![f'(c) = \frac{[(p-q)z^{p+q-1}+pz^{p-1}-qz^{q-1}]z' + z^q - z^p}{(z^q + c)^2}](/cgi-bin/mimetex.cgi? f'(c) = \frac{[(p-q)z^{p+q-1}+pz^{p-1}-qz^{q-1}]z' + z^q - z^p}{(z^q + c)^2} )
The formula for f'(c) is needed if you want to apply the distance estimator method (here z' is the value of the derivative at the previous iteration).
The formula for f'(z) shows that 0 is always a critical point, but the other critical points depend on c and need to solve a polynomial.
I assume that the pictures are for the critical point zero ?
I have tried to find a formula for the derivatives, assuming p > q > 1 :
The formula for f'(c) is needed if you want to apply the distance estimator method (here z' is the value of the derivative at the previous iteration).
The formula for f'(z) shows that 0 is always a critical point, but the other critical points depend on c and need to solve a polynomial.
I assume that the pictures are for the critical point zero ?
Title: Re: Strange Mandelbrot Set
Post by: element90 on June 11, 2014, 12:44:40 PM
Quote
I assume that the pictures are for the critical point zero ?
Finding the critical points using f'(z) = 0 for the first formula results in a critical point at zero and the other 10 critical points are dependent on the position in the complex plane. However, the critical point at zero is useless because after the first iteration all subsequent iterations result in 1.
I've checked, those pictures use c as the initial value of z so result in 'perturbed' fractals as c (the location in the complex plane) is not a critical point.
I've added a variation of these formulae to Saturn.
z = (z^alpha + beta)/(z^gamma + delta) + epsilon + zeta
with the following default values:
alpha = 10
beta = position in the complex plane
gamma = 8
delta = position in the complex plane
epsilon = position in the complex plane
zeta = -2
z0 = 0
Pictures soon ...
At some point formulae of this type will be added to Neptune (my multiple critical point fractal program), I have to implement critical points dependent on the location in the complex plane. The critical points will have to be evaluated for all locations instead of once.
Title: Re: Strange Mandelbrot Set
Post by: element90 on June 11, 2014, 07:24:11 PM
As promised pictures using this formula:
z = (z^alpha + beta)/(z^gamma + delta) + epsilon + zeta
called "Mandelbrot Division" in the development version of Saturn and Titan.
(https://copy.com/i2fXfJpsi8Sx)
https://copy.com/i2fXfJpsi8Sx (https://copy.com/i2fXfJpsi8Sx)
(https://copy.com/KbokiLLRNZV9)
https://copy.com/KbokiLLRNZV9 (https://copy.com/KbokiLLRNZV9)
(https://copy.com/2i1WpFjehFrx)
https://copy.com/2i1WpFjehFrx (https://copy.com/2i1WpFjehFrx)
(https://copy.com/GCLyGBzkaxzn)
https://copy.com/GCLyGBzkaxzn (https://copy.com/GCLyGBzkaxzn)
(https://copy.com/d8ArqmPLoVT0)
https://copy.com/d8ArqmPLoVT0 (https://copy.com/d8ArqmPLoVT0)
(https://copy.com/NC1aLSDZmCq9)
https://copy.com/NC1aLSDZmCq9 (https://copy.com/NC1aLSDZmCq9)
(https://copy.com/ATF8y3kyG4TL)
https://copy.com/ATF8y3kyG4TL (https://copy.com/ATF8y3kyG4TL)
(https://copy.com/AszlNwGvJuDq)
https://copy.com/AszlNwGvJuDq (https://copy.com/AszlNwGvJuDq)
(https://copy.com/jlVGSIIdBxiK)
https://copy.com/jlVGSIIdBxiK (https://copy.com/jlVGSIIdBxiK)
(https://copy.com/qu11GLfWVXph)
https://copy.com/qu11GLfWVXph (https://copy.com/qu11GLfWVXph)
z = (z^alpha + beta)/(z^gamma + delta) + epsilon + zeta
called "Mandelbrot Division" in the development version of Saturn and Titan.
(https://copy.com/i2fXfJpsi8Sx)
https://copy.com/i2fXfJpsi8Sx (https://copy.com/i2fXfJpsi8Sx)
(https://copy.com/KbokiLLRNZV9)
https://copy.com/KbokiLLRNZV9 (https://copy.com/KbokiLLRNZV9)
(https://copy.com/2i1WpFjehFrx)
https://copy.com/2i1WpFjehFrx (https://copy.com/2i1WpFjehFrx)
(https://copy.com/GCLyGBzkaxzn)
https://copy.com/GCLyGBzkaxzn (https://copy.com/GCLyGBzkaxzn)
(https://copy.com/d8ArqmPLoVT0)
https://copy.com/d8ArqmPLoVT0 (https://copy.com/d8ArqmPLoVT0)
(https://copy.com/NC1aLSDZmCq9)
https://copy.com/NC1aLSDZmCq9 (https://copy.com/NC1aLSDZmCq9)
(https://copy.com/ATF8y3kyG4TL)
https://copy.com/ATF8y3kyG4TL (https://copy.com/ATF8y3kyG4TL)
(https://copy.com/AszlNwGvJuDq)
https://copy.com/AszlNwGvJuDq (https://copy.com/AszlNwGvJuDq)
(https://copy.com/jlVGSIIdBxiK)
https://copy.com/jlVGSIIdBxiK (https://copy.com/jlVGSIIdBxiK)
(https://copy.com/qu11GLfWVXph)
https://copy.com/qu11GLfWVXph (https://copy.com/qu11GLfWVXph)
Title: Re: Strange Mandelbrot Set
Post by: jdebord on June 12, 2014, 09:23:45 AM
Thank you element90 for the explanations and of course the wonderful pictures !
BTW, my previous formula for the iterated derivative f'(c) was erroneous. It has been corrected.
BTW, my previous formula for the iterated derivative f'(c) was erroneous. It has been corrected.
Title: Re: Strange Mandelbrot Set
Post by: David Makin on June 12, 2014, 11:38:36 PM
With respect to critical points - an idea I had for a different Newton formula is probably transferable - in that I set the user params so they specified the roots and the formula constructed the Newton to be iterated accordingly.
Obviously one could do the same by allowing the user to specify critical points as user parameters - this would also give an interesting way for user-controlled morphing (as with my Newton formula).
Obviously one could do the same by allowing the user to specify critical points as user parameters - this would also give an interesting way for user-controlled morphing (as with my Newton formula).